If $(1+x)^{2016} = a_{0}+a_{1}x + a_{2}x^{2} + \cdots + a_{2016}x^{2016} \ \ \ $ when $a_{n}$ is coefficient of $x^{n}$
Then , find the value of $\ \ \ \frac{a_{0}}{1}-\frac{a_{1}}{5} + \frac{a_{2}}{9} - \frac{a_{3}}{13}+\cdots +\frac{a_{2016}}{8065} $
My note: $\left (\frac{a_{0}}{1} + \frac{a_{2}}{9} +\frac{a_{4}}{17} + \cdots + \frac{a_{2016}}{8065} \right ) - \left (\frac{a_{1}}{5} + \frac{a_{3}}{13} +\frac{a_{5}}{21} + \cdots + \frac{a_{2015}}{8061} \right )$
$ = \sum_{n=0}^{1008}\frac{\binom{2016}{2n}}{8n+1} - \sum_{n=0}^{1007}\frac{\binom{2016}{2n+1}}{8n+5}$ $\ \ \ \ = \frac{1}{8065} + \sum_{n=0}^{1007} \frac{2016!}{\left ( 2016-2n \right )!\left ( 2n+1 \right )!} \left ( \frac{2n+1}{8n+1} - \frac{2016-2n}{8n+5}\right ) $
I got stuck to this problem all day , I can't rearrange the final above to collapse them.
Thank you in advance , all of your solution light me up for keep practicing mathematics.
@Ivan Neretin is right. You have $$ \text{Your sum}=\int_0^1(1-x^4)^{2016}dx=\prod_{j=1}^{2016}\left(1-\frac1{4j}\right) =\frac{\Gamma \left(2016\frac{3}{4}\right)}{\Gamma \left(\frac{3}{4}\right) \Gamma (2017)}. $$